Question 3.2: Find the bus impedance matrix of Example 3.1 by inversion an......
Find the bus impedance matrix of Example 3.1 by inversion and also by open-circuit testing. The inversion of the admittance matrix in Equation 3.16,
= \left|\begin{matrix} 1-j5 & j5 & 0 \\ j5 & 0.5 – j8.33 & j3.33 \\ 0 & j3.33 & 0.33-j3.33 \end{matrix} \right| (3.16)
calculated in Example 3.1, gives
\overline{Z}_B =\left|\begin{matrix} 0.533+j0.05 & 0.543-j0.039 & 0.533-j0.0.92 \\ 0.543-j0.039 & 0.55+j0.069 & 0.552+j0.014 \\ 0.533-j0.092 & 0.522+j0.014 & 0.577+j0.258 \end{matrix} \right| (3.20)
From Equation 3.20 we note that the zero elements of the bus admittance matrix get populated in the bus impedance matrix. As we will discover, the admittance matrix for power system networks is sparse and this sparsity is lost in the impedance matrix.
The same bus impedance matrix can be constructed from the open-circuit test results. Unit currents are injected, one at a time, at each bus and the other current sources are open circuited. Figure 3.3c shows unit current injected at bus 1 with bus 3 current source open circuited. Z_{11} is given by voltage at node 1 divided by current I_{1} (=1.0 per unit):
1\left\|[(2 + j0.02)\right\|(3 + j0.03)] = Z_{11} = 0.533 + j0.05
The injected current divides as shown in Figure 3.3c. Transfer impedance Z_{12} = Z_{21} at bus 2 is the potential at bus 2. Similarly, the potential at bus 3 gives Z_{13} = Z_{31}. The example shows that this method of formation of bus impedance matrix is tedious.
